Rocket acceleration problem: confused about Newton's 2nd Law

In summary, the first approach is incorrect because the mass of the object cannot change. The second approach is correct, but requires treating the ejecta as an external force.
  • #1
8
0
Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: I approach a rocket acceleration problem using two approaches: F=d(m*v)/dt and F=ma. The resulting differential equations are different. What am I doing wrong?

We have a ship with a mass-reaction rocket engine floating in space.
The initial mass of the ship (including fuel) is m0 [kg].
The rocket produces a constant thrust F [N].
The burn rate of the fuel is R [kg/s].
The intitial speed v of the ship is 0 m/s.

What is the velocity/acceleration of the ship?

I first approached this problem like this:

F = d(m*v)/dt (using the momentum version of Newton's 2nd Law)
F = m dv/dt + v dm/dt
F = (m0 - R t) dv/dt + v(-R)
dv/dt = (F + R v) / (m0 - R t)

OK, so it's a differential equation for v.
Next, I approached it like this:

F = m a (the more commonly encountered version of Newton's 2nd Law)
F = (m0 - R t) a
a = F / (m0 - R t)

but a is also dv/dt, isn't it? so I get

dv/dt = F / (m0 - R t)

Compare this with dv/dt from the first approach. They're different. I'm missing a whole term Rv/(m0-Rt).
What am I doing wrong?

Thanks for reading this far.
 
Physics news on Phys.org
  • #2
aniseed said:
TL;DR Summary: I approach a rocket acceleration problem using two approaches: F=d(m*v)/dt and F=ma. The resulting differential equations are different. What am I doing wrong?

We have a ship with a mass-reaction rocket engine floating in space.
The initial mass of the ship (including fuel) is m0 [kg].
The rocket produces a constant thrust F [N].
The burn rate of the fuel is R [kg/s].
The intitial speed v of the ship is 0 m/s.

What is the velocity/acceleration of the ship?

I first approached this problem like this:

F = d(m*v)/dt (using the momentum version of Newton's 2nd Law)
F = m dv/dt + v dm/dt
This is wrong. There are several threads on here about it. The mass of an object cannot change. Instead, you may redefine the total mass that makes up an object.

Consider a truck leaking sand. The mass of the truck plus remaining sand decreases, hence the total momentum of what's left decreases, but there is no retarding force involved.
aniseed said:
OK, so it's a differential equation for v.
Next, I approached it like this:

F = m a (the more commonly encountered version of Newton's 2nd Law)
F = (m0 - R t) a
a = F / (m0 - R t)

but a is also dv/dt, isn't it? so I get

dv/dt = F / (m0 - R t)
If there is a constant force applied to what remains defined as the rocket, then this must be correct.
 
  • Like
Likes erobz
  • #3
  • Like
Likes hutchphd, MatinSAR, erobz and 1 other person
  • #4
aniseed said:
What am I doing wrong?
Konstantin Tsiolkovsky approves your second approach, which is always a plus in rocketry.

I think the problem with your first approach is that, while you are free to apply the chain rule like that, doing that is not consistent with the way you've defined force in the question. PeroK's example of a truck leaking sand is good, but a slightly better one in this context is a pair of equally powerful thrusters on opposite sides of the ship firing. The result is obviously no acceleration, but there's also obviously a negative ##dm/dt## so your first equation would say there's a net applied force. That's not the customary usage of "force" and it's not consistent with the "rocket applies a constant force" that you mean here.
 
Last edited:
  • Like
Likes erobz and vanhees71
  • #5
In the second approach you need to treat the ejecta as applying an external force (Thrust) to the rocket which has instantaneous mass ##m##. You determine what that force is by the derivation linked in post #3. So neither of your approaches hit the mark, but the second one is salvageable.
 
  • Like
Likes Ibix
  • #6
The root of your problem is that you wrote F = ma and not Fnet = ma and treated this F as an external force on the rocket. Here is a straightforward derivation that uses Newton's second law and considers the rocket and its contents as an isolated system.

The net force on the system, defined as the rocket plus all the fuel, is zero. At any time ##t##, the momentum of the system, is the sum of the momenta of the rocket, the unburnt fuel and the ejecta. However the momentum of the ejecta does not change once ejected. Thus, at any time ##t##,$\begin{align}& 0=\frac{dp_{\text{sys.}}}{dt}=\frac{d}{dt}\left(p_{\text{rocket}}+p_{\text{unb. fuel}}\cancel{+p_{\text{ejecta}}}\right)=m\frac{dv}{dt}+v\frac{dm}{dt}=(m_0-Rt)\frac{dv}{dt}-Rv\nonumber \\ & \implies \frac{dv}{v}=R\frac{dt}{m_0-Rt}.\nonumber \end{align}$Here ##m_0## is the mass of the system at ##t=0## and ##m(t)=m_0-Rt## is the mass of the rocket plus the unburnt fuel.


Sorry about the mess.
 
Last edited:
  • #7
kuruman said:
The root of your problem is that you wrote F = ma and not Fnet = ma and treated this F as an external force on the rocket. Here is a straightforward derivation that uses Newton's second law and considers the rocket and its contents as an isolated system.

The net force on the system, defined as the rocket plus all the fuel, is zero. At any time ##t##, the momentum of the system, is the sum of the momenta of the rocket, the unburnt fuel and the ejecta. However the momentum of the ejecta does not change once ejected. Thus, at any time ##t##,$$\begin{align}& 0=\frac{dp_{\text{sys.}}}{dt}=\frac{d}{dt}\left(p_{\text{rocket}}+p_{\text{unb. fuel}}\cancel{+p_{\text{ejecta}}}\right)=m\frac{dv}{dt}+v\frac{dm}{dt}=(m_0-Rt)\frac{dv}{dt}-Rv\nonumber \\ & \implies \frac{dv}{v}=R\frac{dt}{m_0-Rt}.\nonumber \end{align}$$Here ##m_0## is the mass of the system at ##t=0## and ##m(t)=m_0-Rt## is the mass of the rocket plus the unburnt fuel.
But that doesn't seem to be the rocket equation?

$$\sum F = M \dot v_r + \dot M v_{e/r}$$
 
  • #8
$$ m\frac{dv}{dt}+v\frac{dm}{dt}=(m_0-Rt)\frac{dv}{dt}-Rv$$

How can they be equivalent on the RHS?

This first terms are equivalent:
$$(m_0-Rt)\frac{dv}{dt} = M \dot v_r$$

The second terms don't seem to be?

$$ \dot M v_{e/r} \neq \dot M v_r$$
 
  • #9
erobz said:
How can they be equivalent on the RHS?

This first terms are equivalent:The second terms don't seem to be?

$$ \dot M v_{e/r} \neq \dot M v_r$$
I may have messed up the momentum change of the ejecta. Let me rethink this.
 
  • Like
Likes scottdave and erobz
  • #10
W.r.t. a stationary frame ##O## the ejecta is being carried along at ##\vec{v}_{r/O}## at time ##t##, then at ##t+ \Delta t## it's moving at ##\vec{v}_{e/O}##. I think the ejecta has a non-zero rate of change of momentum?
 
  • #11
How about this?

$$\sum F_{ext} = \frac{d}{dt} \left( p_{\rm{rocket}} + p_{\rm{fuel}} + p_{\rm{ejecta}} \right)$$

Mass of the rocket is constant ##M_r##

The mass of fuel lost from the rocket is gained by the ejecta ## -\frac{dM_f}{dt} = \frac{dM_e}{dt}##

The velocity of the ejecta in the rest frame is ##v_{e/O} = v_r - v_{e/r}## and is constant, but the mass of the ejecta is accumulating at rate ##\frac{dM_e}{dt}##.

$$\sum F_{ext} = \overbrace{\left( M_r\frac{dv_r}{dt} \right)}^{\rm {rocket}} +\overbrace{\left(M_f\frac{dv_r}{dt} + \frac{dM_f}{dt}v_r\right)}^{\rm{fuel}}+ \overbrace{ \left( \frac{dM_e}{dt} \left( v_r - v_{e/r}\right) + \cancel{ M_e \frac{d}{dt} ( v_r -v_{e/r} ) }^0 \right)}^{\rm{ejecta}} $$

$$\sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt}+ \cancel{ \left( \frac{dM_f}{dt}v_r + \frac{dM_e}{dt}v_r \right)}^0 -\frac{dM_e}{dt} v_{e/r}$$

$$ \sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt} + \frac{dM_f}{dt} v_{e/r}$$

That seems to be it? But did I get there by hook or by crook?
 
Last edited:
  • #12
erobz said:
W.r.t. a stationary frame ##O## the ejecta is being carried along at ##\vec{v}_{r/O}## at time ##t##, then at ##t+ \Delta t## it's moving at ##\vec{v}_{e/O}##. I think the ejecta has a non-zero rate of change of momentum?
Yes, that's what I messed up. I don't think it can be done without Newton's third law. The system as I had it is isolated and combustion adds energy but not momentum.
 
  • #13
erobz said:
The velocity of the ejecta in the rest frame is ve/O=vr−ve/r and is constant,
I thought it was ##v_{e/r}## that was constant.
 
  • #14
haruspex said:
I thought it was ##v_{e/r}## that was constant.
I think ##v_{e/r}## could be anything. The reason why I say that ##v_{e/O}## is constant is because once it leaves the rocket it's not accelerating in the inertial frame. So the "packet of ejecta" that changed the rockets momentum is moving at a constant velocity w.r.t ground. but, not the same velocity as the other packets. Is that making sense? The mass of ejecta is accumulating, but the momentum of currently expelled mass is constant in time.

That is my interpretation of the following statement in Hibbeler:
InkedIMG_1791.jpg
 
Last edited:
  • #15
erobz said:
I think ##v_{e/r}## could be anything. The reason why I say that ##v_{e/O}## is constant is because once it leaves the rocket it's not accelerating in the inertial frame. So the "packet of ejecta" that changed the rockets momentum is moving at a constant velocity w.r.t ground.
But when you write ##p_{ejecta}##, that is the momentum of the ejecta as a whole, so how are you defining the velocity of the ejecta as a whole? The only obvious way is as the velocity of its mass centre, and that is not constant.
If you skip consideration of the ejecta's velocity and just address its change in momentum, clearly that is its added mass multiplied by the rest frame velocity of that after ejection, as desired.
 
  • #16
haruspex said:
But when you write ##p_{ejecta}##, that is the momentum of the ejecta as a whole, so how are you defining the velocity of the ejecta as a whole? The only obvious way is as the velocity of its mass centre, and that is not constant.
If you skip consideration of the ejecta's velocity and just address its change in momentum, clearly that is its added mass multiplied by the rest frame velocity of that after ejection, as desired.
I don’t know. Probably got there by coincidence then. I’m guessing you aren’t seeing a path forward in the that style. Anyhow, the impulse/momentum is easy enough to derive. I was just seeing if it was salvageable.
 
  • #17
I wonder if the book derivation is incorrect too for the reason you mention? They say in italics the control volume includes both the mass of the device and the expelled mass. It should fall prey to the same argument you are making?
 
  • #18
erobz said:
I wonder if the book derivation is incorrect too for the reason you mention? They say in italics the control volume includes both the mass of the device and the expelled mass. It should fall prey to the same argument you are making?
No, the book argument is fine. It does not create a variable to represent the velocity of what has already been ejected, just notes that its momentum doesn’t change. It should be possible to make what you wrote work with a careful definition of that velocity, but why bother?
 
  • #19
haruspex said:
It should be possible to make what you wrote work with a careful definition of that velocity, but why bother?
Agreed, it was just a curiosity spurred by what was posted.
 
  • #20
aniseed said:
Compare this with dv/dt from the first approach. They're different. I'm missing a whole term Rv/(m0-Rt).
What am I doing wrong?

Thanks for reading this far.
At time t
[tex]P=m(t)v(t)[/tex]
At time t + ##\triangle t##
[tex]P=m(t+\triangle t) v(t+\triangle t) + R\triangle t \{v(t)+U\}[/tex]
where U is speed of ejected fuel relative to rocket. U < 0,
[tex]m(t)=m(0)-Rt[/tex]
RHS first term is momentum rocket body and remaing fuel hold.
RHS second term is momentum that ejected fuel holds.
Conservation of momentum
[tex]m(t+\triangle t)v (t+\triangle t) - m(t)v(t) + R\triangle t \{v(t)+U\} = 0[/tex]
[tex]\frac{d}{dt}[m(t)v (t)] = - R\{v(t)+U\} = F - Rv(t)[/tex]
Your first way does not correspond to F but F-Rv. From this we get easily
[tex]m(t)\frac{d}{dt}v (t)= F [/tex]
your second way.
 
Last edited:
  • #21
anuttarasammyak said:
At time t
[tex]P=m(t)v(t)[/tex]
At time t + ##\triangle t##
[tex]P=m(t+\triangle t) v(t+\triangle t) + R\triangle t \{v(t)+U\}[/tex]
If ##P(t)=m(t)v(t)## as a general statement then by definition ##P(t+\Delta t)=m(t+\triangle t) v(t+\triangle t)##.
Or do you mean ##P(t)=m(t+\triangle t) v(t+\triangle t)+ R\triangle t \{v(t)+U\}##?

Anyway, what the OP wants to know is specifically where s/he went wrong in post #1.
 
  • #22
haruspex said:
If P(t)=m(t)v(t) as a general statement then by definition P(t+Δt)=m(t+△t)v(t+△t).
No, I meant P is just a momentum value that the system t and the system t+##\triangle## t share by momentum conservation law.
 
Last edited:
  • #23
anuttarasammyak said:
No, I meant P is just a momentum value that the system t and the system t+##\triangle## t share by momentum conservation law.
Then you need to define your system. What exactly does it consist of?
 
  • #24
The system consists of Rocket body and fuel. All the fuel is contained in rocket at time t. Some fuel are ejected from Rocket during t and t+##\triangle## t. All the fuel ejected before time t, which continues inertial motion of constant mometum any way, does not matter here. If we choose t=0 Rocket has full tank.
 
Last edited:
  • #25
haruspex said:
Anyway, what the OP wants to know is specifically where s/he went wrong in post #1.
m(t) is defined by man made convention of removing ejected fuel mass so d/dt [m(t)v(t)] ##\neq## F does not surprise me. For simplicity say F=0, fuel are just disposed with no thrust or kept in tank but we take off account for m(t). From my post #20
[tex]\frac{d}{dt}[m(t)v(t)]+Rv(t)=0[/tex]
We are counting less and less mass for m(t) as
[tex]m(t)=m(0)-Rt[/tex]
so its momentum decreases even in inertial motion v(t)=const. Rv(t) is momentum (per second) of fuel that we are disposing or taking off account. They cancel of course.
[tex]m(t) \frac{d\ v(t)}{dt}=F[/tex] holds because the convention does not matter for a moment-fixed m(t). I hope this answers the questions of OP.
 
Last edited:
  • #26
anuttarasammyak said:
The system consists of Rocket body and fuel.
Still a bit vague. From "P=m(t)v(t)", it seems to be the rocket and unburnt fuel at some particular time t, so really it is, as I suggested, P(t), and ##P(t)=m(t+\triangle t) v(t+\triangle t)+ R\triangle t \{v(t)+U\}##.
From which I deduce that m(t) is the rocket + unburnt fuel at a variable time t, so not the same system.
Yes, that all works, but it does need the variables to be defined.
 
  • #27
Why not simply using the fundamental law of momentum conservation? For simplicity let's look at the motion in one direction ("vertically up").

In an infinitesimal time interval ##(t,t+\mathrm{d} t)##. Let ##\vec{p}=m \vec{v}## be the momentum of the rocket and ##v_e## the velocity of the fuel leaving the rocket ("ejecta") as measured in the restframe of the rocket. For simplicity we assume that ##v_e=\text{const}##. Then the change of momentum of the rocket is
$$p(t+\mathrm{d} t)-p(t)=\mathrm{d} t \dot{p}=\mathrm{d} t [\dot{m} v + m \dot{v}]=\mathrm{d} t(m \dot{v}-\mu v)=\mathrm{d} t [-m g - \mu(v-v_e)]$$
On the left-hand side we have the change of the momentum of the rocket, which is due to the gravitational force of the Earth acting on it and the change of momentum due to the exhausted fuel with the mass ##\mathrm{d}m_{\text{fuel}}=-\mathrm{d} t \mu##.
Finally we get
$$m \dot{v} = -m g +\mu v_e=-mg-\dot{m} v_e.$$
Assuming that also ##\dot{m}=-\mu=\text{const}## we have ##m=m_0-\mu t## and thus
$$(m_0-\mu t) \dot{v}=-(m_0-\mu t) g + \mu v_e$$
or
$$\dot{v}=-g + \frac{\mu}{m_0-\mu t} v_e.$$
This we can simply integrate
$$v(t)=v_0 - g t -v_e \ln \left (\frac{m_0-\mu t}{m_0} \right).$$
 
  • #28
vanhees71 said:
Why not simply using the fundamental law of momentum conservation?
Fine, but the question the OP asks is what specifically is wrong with the failed method in post #1.
 
  • #29
haruspex said:
Yes, that all works, but it does need the variables to be defined.
More comprehensive treatment. Say rocket is at rest at t=0. Momentum of the system is conserved zero.
[tex]0=m(t)v(t) + R \int_0^t ( v(u)+U )du[/tex]
differentiating by t,
[tex]\frac{d}{dt}[m(t)v(t)]+R [ v(t)+U ]=\frac{d}{dt}[m(t)v(t)]+Rv(t)-F=0[/tex]
where m(t) is mass of rocket body and fuel in tank. ##m(t)=m(0)-Rt##,
##v(t)## is speed of rocket,
##U## is relative speed of ejected fuel to rocket. ##U<0## when ##v(t)>0##,
##F=-RU##.
 
Last edited:
  • Like
Likes erobz
  • #30
haruspex said:
Fine, but the question the OP asks is what specifically is wrong with the failed method in post #1.
Obviously the momentum carried by the exhaust hasn't been taken into account. Isn't this obvious from the correct derivation?
 
  • #31
anuttarasammyak said:
More comprehensive treatment. Say rocket is at rest at t=0. Momentum of the system is conserved zero.
[tex]0=m(t)v(t) + R \int_0^t ( v(u)+U )dt[/tex]
differentiating by t,
[tex]\frac{d}{dt}[m(t)v(t)]+R [ v(t)+U ]=\frac{d}{dt}[m(t)v(t)]+Rv(t)-F=0[/tex]
where m(t) is mass of rocket body and fuel in tank. ##m(t)=m(0)-Rt##,
##v(t)## is speed of rocket,
##U## is relative speed of ejected fuel to rocket. ##U<0## when ##v(t)>0##,
##F=-RU##.
So, since you’ve defined the ejecta momentum to be that integral, that treatment works. So it wasn’t all that difficult to correct after all. Thanks for sharing!

P.S. in your top line you’ve wrote ##v(u)## in the integral.
 
  • Like
Likes vanhees71 and anuttarasammyak
  • #32
erobz said:
P.S. in your top line you’ve wrote ##v(u)## in the integral.
Yes, it is in the integral. My typo not dt but du, dummy variable for integration. I corrected it in the post. Thanks.
 
  • Like
Likes vanhees71 and erobz
  • #33
@anuttarasammyak So, even more generally it starts out with:

$$ 0 = m(t)v(t) - \int_{0}^{t} \dot m(u) ( v(u)+U(u))~du $$

I think the momentum of the rocket + unburnt fuel should always be opposite the momentum of the ejecta

but since ##\dot m ## and ## v + U ## are both negative, we need ##m(t)v(t)## to be positive. Which is leaning me toward the minus sign?
 
Last edited:
  • #34
Just for clarity, correcting post #11 using the proper way to capture the accumulation of momentum in the ejecta from @anuttarasammyak:

$$\sum F_{ext} = \frac{d}{dt} \left( p_{\rm{rocket}} + p_{\rm{fuel}} + p_{\rm{ejecta}} \right)$$

Mass of the rocket is constant ##M_r##

The mass of fuel lost from the rocket is gained by the ejecta ## \frac{dM_e}{dt} = -\frac{dM_f}{dt}##

The velocity of the ejecta in the rest frame is ##v_{e/O} = v_r - v_{e/r}## and is constant after being ejected, but the momentum of the ejecta is accumulating, hence:

$$\sum F_{ext} = \frac{d}{dt} \left( M_r v_r + M_f v_r + \int \frac{dM_e}{dt}( v_r - v_{e/r}) ~dt \right)$$

Applying the derivative:

$$\sum F_{ext} = \overbrace{\left( M_r\frac{dv_r}{dt} \right)}^{\rm {rocket}} +\overbrace{\left(M_f\frac{dv_r}{dt} + \frac{dM_f}{dt}v_r\right)}^{\rm{fuel}}+ \overbrace{ \frac{dM_e}{dt}( v_r - v_{e/r})}^{\rm{ejecta}} $$

$$\sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt}+ \cancel{ \left( \frac{dM_f}{dt}v_r + \frac{dM_e}{dt}v_r \right)}^0 -\frac{dM_e}{dt} v_{e/r}$$

The Rocket Equation:

$$ \sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt} + \frac{dM_f}{dt} v_{e/r}$$
 
Last edited:
  • Like
Likes vanhees71 and kuruman
  • #35
vanhees71 said:
Obviously the momentum carried by the exhaust hasn't been taken into account. Isn't this obvious from the correct derivation?
To the OP, almost surely not. But your simple explanation quoted above, together with the example in post #2, should do it.
 

Suggested for: Rocket acceleration problem: confused about Newton's 2nd Law

Back
Top